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what tricks unlock this integral?? 

Michael Penn
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1 окт 2024

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Комментарии : 25   
@jeffreywood4759
@jeffreywood4759 13 часов назад
<a href="#" class="seekto" data-time="560">9:20</a> For those wondering, use the substitution u = (1-t) for the second part of the integral, you swap the bounds of integration, but pick up a minus sign from the derivative, so the minus becomes positive and the bounds swap back to 0 to 1. Then just do the trivial substitution of u -> t, and you'll see the two parts of the integral are the same.
@unturnedd
@unturnedd 12 часов назад
could have mentioned in the end that the result is equivalent to pi/sinh(pi)
@karnetik
@karnetik 13 часов назад
i see a michael penn integral, i click
@ruilopes6638
@ruilopes6638 12 часов назад
Deep Michael Penn lore on this video. Many recalls
@goodplacetostop2973
@goodplacetostop2973 14 часов назад
<a href="#" class="seekto" data-time="775">12:55</a>
@jesusthroughmary
@jesusthroughmary 14 часов назад
And here I thought I was first
@benardolivier6624
@benardolivier6624 13 часов назад
It was a good place to st... 😉
@dang-x3n0t1ct
@dang-x3n0t1ct 14 часов назад
Why not use cosx= Re(e^ix) ?
@xizar0rg
@xizar0rg 13 часов назад
Because that wouldn't show the same techniques he's demonstrating here. It's a pedagogic choice rather than a pragmatic one.
@eiseks3410
@eiseks3410 12 часов назад
I agree, it was an overkill
@danielc.martin
@danielc.martin 9 часов назад
Better
@59de44955ebd
@59de44955ebd 8 часов назад
Footnote: the final result can also be expressed as pi * csch(pi), where csch is the hyperbolic cosecant
@alipourzand6499
@alipourzand6499 13 часов назад
At some point, I was affraid to see a complex result for the integral, but there was a happy end!
@wolfmanjacksaid
@wolfmanjacksaid 13 часов назад
<a href="#" class="seekto" data-time="207">3:27</a> not a big fan of setting "u" equal to two different things
@karnetik
@karnetik 13 часов назад
u=1/u😵
@noxfortes
@noxfortes 13 часов назад
Also noticed it. Should've used 'v', maybe.
@lythd
@lythd 12 часов назад
its color coded and with an arrow specifying which is for which integral. once on the inside of integral they are the same thing, an integration variable over the specified bounds. so it doesn't really matter how they related to the previous integration variable. so its not really two different things, it just relates to the previous integration variable differently in each integral, and its very clearly defined which is which
@ClaudioCabrera-d4g
@ClaudioCabrera-d4g 11 часов назад
Hi Michael, Claudio from Chile
@jardozouille1677
@jardozouille1677 8 часов назад
So complicated ... wouldn't it be interesting to start by noting that ln(x/(2-x)) = 2 argth(x-1) ?
@angelosettanni559
@angelosettanni559 8 часов назад
i can' t understand the final result. can someone help me to understand?
@ClaudioCabrera-d4g
@ClaudioCabrera-d4g 11 часов назад
How can I send you a similar problem so that you have a try in the channel?
@BrunoMoreiraTorres
@BrunoMoreiraTorres 14 часов назад
Funny things always seem to happen when you get an integral with bounds at 0, 1 or infinity and applies u=1/x 🤔
@meurdesoifphilippe5405
@meurdesoifphilippe5405 13 часов назад
I love this world where there is no definition problem in 0
@士-x7e
@士-x7e 8 часов назад
π/Sinh π
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